Answer :
Final Answer:
The probability that a single randomly selected value is greater than 43.5 is approximately 0.2119. The probability that a sample of size 119 is randomly selected with a mean greater than 43.5 is approximately 0.0000.
Explanation:
Given a normal distribution with a mean [tex]\(\mu = 15\)[/tex] and a standard deviation \[tex](\sigma = 97.3\),[/tex]we are asked to find two probabilities:
Probability for a Single Value:
To find[tex]\(P(X > 43.5)\), where \(X\)[/tex] represents a single randomly selected value, we calculate the z-score corresponding to[tex]\(x = 43.5\):\[z = \frac{43.5 - 15}{97.3} \approx 0.2965\][/tex]
Using the standard normal distribution table or calculator, we find that the probability \[tex](P(X > 43.5)\)[/tex]is approximately 0.2119.
Probability for Sample Mean
To find[tex]\(P(M > 43.5)\),[/tex]where[tex]\(M\)[/tex]represents the mean of a sample of size \(n = 119\), we calculate the standard error of the mean (\(SE_M\)):[tex]\[SE_M = \frac{\sigma}{\sqrt{n}} = \frac{97.3}{\sqrt{119}} \approx 8.8967\][/tex]
Then, we calculate the z-score for[tex]\(x = 43.5\)[/tex] using the formula:
[tex]\[z = \frac{43.5 - 15}{SE_M} \approx 3.1913\][/tex]
Using the standard normal distribution table or calculator, we find that the probability[tex]\(P(M > 43.5)\[/tex]) is approximately 0.0000.
In summary, for part (a), the probability that a single randomly selected value is greater than 43.5 is approximately 0.2119. For part (b), the probability that a sample of size 119 is randomly selected with a mean greater than 43.5 is approximately 0.0000.
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